# What is the vertex form of y= 9x^2 + 2x + 2/7 ?

Feb 26, 2018

See below:

#### Explanation:

$y = a {\left(x - h\right)}^{2} + k$ with $\left(h , k\right)$ as the vertex.

To find the vertex form of a quadratic equation, complete the square:

$y = 9 \left({x}^{2} + \frac{2}{9} x + {\left(\frac{1}{9}\right)}^{2} - {\left(\frac{1}{9}\right)}^{2}\right) + \frac{2}{7}$

$y = 9 {\left(x + \frac{1}{9}\right)}^{2} - \frac{9}{81} + \frac{2}{7}$

$y = 9 {\left(x + \frac{1}{9}\right)}^{2} + \frac{11}{63}$

The vertex is $\left(- \frac{1}{9} , \frac{11}{63}\right)$

You can also find the vertex with formulas:

$h = - \frac{b}{2 a}$

$k = c - {b}^{2} / \left(4 a\right)$

$- - - - - - - - - - - -$

$h = - \frac{2}{2 \cdot 9} = - \frac{1}{9}$

$k = \frac{2}{7} - {\left(- 2\right)}^{2} / \left(4 \cdot 9\right) = \frac{2}{7} - \frac{4}{36} = \frac{11}{63}$

so the vertex is at

$\left(- \frac{1}{9} , \frac{11}{63}\right)$

You can also find vertex form this way:

$y = a \left(x + \frac{1}{9}\right) + \frac{11}{63}$

Plug in $a$ from the original equation:

$y = 9 \left(x + \frac{1}{9}\right) + \frac{11}{63}$

Apologies for the length :)